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4x^2-35x+48=-6x+3
We move all terms to the left:
4x^2-35x+48-(-6x+3)=0
We get rid of parentheses
4x^2-35x+6x-3+48=0
We add all the numbers together, and all the variables
4x^2-29x+45=0
a = 4; b = -29; c = +45;
Δ = b2-4ac
Δ = -292-4·4·45
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-11}{2*4}=\frac{18}{8} =2+1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+11}{2*4}=\frac{40}{8} =5 $
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